3.31.70 \(\int \frac {-5+x \log (x)+(6 x-2 x^2+(-5 x-4 x^2) \log (x)) \log (2 x)+(x+x \log (x)) \log (2 x) \log (\log (2 x))}{(25 x+11 x^2+(-5 x^2-2 x^3) \log (x)) \log (2 x)+(-5 x+x^2 \log (x)) \log (2 x) \log (\log (2 x))} \, dx\)

Optimal. Leaf size=23 \[ \log (x+(5-x \log (x)) (5+2 x-\log (\log (2 x)))) \]

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Rubi [F]  time = 12.52, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-5+x \log (x)+\left (6 x-2 x^2+\left (-5 x-4 x^2\right ) \log (x)\right ) \log (2 x)+(x+x \log (x)) \log (2 x) \log (\log (2 x))}{\left (25 x+11 x^2+\left (-5 x^2-2 x^3\right ) \log (x)\right ) \log (2 x)+\left (-5 x+x^2 \log (x)\right ) \log (2 x) \log (\log (2 x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-5 + x*Log[x] + (6*x - 2*x^2 + (-5*x - 4*x^2)*Log[x])*Log[2*x] + (x + x*Log[x])*Log[2*x]*Log[Log[2*x]])/(
(25*x + 11*x^2 + (-5*x^2 - 2*x^3)*Log[x])*Log[2*x] + (-5*x + x^2*Log[x])*Log[2*x]*Log[Log[2*x]]),x]

[Out]

Log[5 - x*Log[x]] + 55*Defer[Int][1/((-5 + x*Log[x])*(-25 - 11*x + 5*x*Log[x] + 2*x^2*Log[x] + 5*Log[Log[2*x]]
- x*Log[x]*Log[Log[2*x]])), x] + Defer[Int][x/((-5 + x*Log[x])*(-25 - 11*x + 5*x*Log[x] + 2*x^2*Log[x] + 5*Lo
g[Log[2*x]] - x*Log[x]*Log[Log[2*x]])), x] - 20*Defer[Int][(x*Log[x])/((-5 + x*Log[x])*(-25 - 11*x + 5*x*Log[x
] + 2*x^2*Log[x] + 5*Log[Log[2*x]] - x*Log[x]*Log[Log[2*x]])), x] + 2*Defer[Int][(x^2*Log[x]^2)/((-5 + x*Log[x
])*(-25 - 11*x + 5*x*Log[x] + 2*x^2*Log[x] + 5*Log[Log[2*x]] - x*Log[x]*Log[Log[2*x]])), x] - 25*Defer[Int][1/
(x*(-5 + x*Log[x])*Log[2*x]*(-25 - 11*x + 5*x*Log[x] + 2*x^2*Log[x] + 5*Log[Log[2*x]] - x*Log[x]*Log[Log[2*x]]
)), x] + 10*Defer[Int][Log[x]/((-5 + x*Log[x])*Log[2*x]*(-25 - 11*x + 5*x*Log[x] + 2*x^2*Log[x] + 5*Log[Log[2*
x]] - x*Log[x]*Log[Log[2*x]])), x] - Defer[Int][(x*Log[x]^2)/((-5 + x*Log[x])*Log[2*x]*(-25 - 11*x + 5*x*Log[x
] + 2*x^2*Log[x] + 5*Log[Log[2*x]] - x*Log[x]*Log[Log[2*x]])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-5+x \log (x)+\left (6 x-2 x^2+\left (-5 x-4 x^2\right ) \log (x)\right ) \log (2 x)+(x+x \log (x)) \log (2 x) \log (\log (2 x))}{x \log (2 x) \left (25+11 x-5 x \log (x)-2 x^2 \log (x)-5 \log (\log (2 x))+x \log (x) \log (\log (2 x))\right )} \, dx\\ &=\int \left (\frac {1+\log (x)}{-5+x \log (x)}+\frac {-25+10 x \log (x)-x^2 \log ^2(x)+55 x \log (2 x)+x^2 \log (2 x)-20 x^2 \log (x) \log (2 x)+2 x^3 \log ^2(x) \log (2 x)}{x (-5+x \log (x)) \log (2 x) \left (-25-11 x+5 x \log (x)+2 x^2 \log (x)+5 \log (\log (2 x))-x \log (x) \log (\log (2 x))\right )}\right ) \, dx\\ &=\int \frac {1+\log (x)}{-5+x \log (x)} \, dx+\int \frac {-25+10 x \log (x)-x^2 \log ^2(x)+55 x \log (2 x)+x^2 \log (2 x)-20 x^2 \log (x) \log (2 x)+2 x^3 \log ^2(x) \log (2 x)}{x (-5+x \log (x)) \log (2 x) \left (-25-11 x+5 x \log (x)+2 x^2 \log (x)+5 \log (\log (2 x))-x \log (x) \log (\log (2 x))\right )} \, dx\\ &=\log (5-x \log (x))+\int \left (\frac {55}{(-5+x \log (x)) \left (-25-11 x+5 x \log (x)+2 x^2 \log (x)+5 \log (\log (2 x))-x \log (x) \log (\log (2 x))\right )}+\frac {x}{(-5+x \log (x)) \left (-25-11 x+5 x \log (x)+2 x^2 \log (x)+5 \log (\log (2 x))-x \log (x) \log (\log (2 x))\right )}-\frac {20 x \log (x)}{(-5+x \log (x)) \left (-25-11 x+5 x \log (x)+2 x^2 \log (x)+5 \log (\log (2 x))-x \log (x) \log (\log (2 x))\right )}+\frac {2 x^2 \log ^2(x)}{(-5+x \log (x)) \left (-25-11 x+5 x \log (x)+2 x^2 \log (x)+5 \log (\log (2 x))-x \log (x) \log (\log (2 x))\right )}-\frac {25}{x (-5+x \log (x)) \log (2 x) \left (-25-11 x+5 x \log (x)+2 x^2 \log (x)+5 \log (\log (2 x))-x \log (x) \log (\log (2 x))\right )}+\frac {10 \log (x)}{(-5+x \log (x)) \log (2 x) \left (-25-11 x+5 x \log (x)+2 x^2 \log (x)+5 \log (\log (2 x))-x \log (x) \log (\log (2 x))\right )}-\frac {x \log ^2(x)}{(-5+x \log (x)) \log (2 x) \left (-25-11 x+5 x \log (x)+2 x^2 \log (x)+5 \log (\log (2 x))-x \log (x) \log (\log (2 x))\right )}\right ) \, dx\\ &=\log (5-x \log (x))+2 \int \frac {x^2 \log ^2(x)}{(-5+x \log (x)) \left (-25-11 x+5 x \log (x)+2 x^2 \log (x)+5 \log (\log (2 x))-x \log (x) \log (\log (2 x))\right )} \, dx+10 \int \frac {\log (x)}{(-5+x \log (x)) \log (2 x) \left (-25-11 x+5 x \log (x)+2 x^2 \log (x)+5 \log (\log (2 x))-x \log (x) \log (\log (2 x))\right )} \, dx-20 \int \frac {x \log (x)}{(-5+x \log (x)) \left (-25-11 x+5 x \log (x)+2 x^2 \log (x)+5 \log (\log (2 x))-x \log (x) \log (\log (2 x))\right )} \, dx-25 \int \frac {1}{x (-5+x \log (x)) \log (2 x) \left (-25-11 x+5 x \log (x)+2 x^2 \log (x)+5 \log (\log (2 x))-x \log (x) \log (\log (2 x))\right )} \, dx+55 \int \frac {1}{(-5+x \log (x)) \left (-25-11 x+5 x \log (x)+2 x^2 \log (x)+5 \log (\log (2 x))-x \log (x) \log (\log (2 x))\right )} \, dx+\int \frac {x}{(-5+x \log (x)) \left (-25-11 x+5 x \log (x)+2 x^2 \log (x)+5 \log (\log (2 x))-x \log (x) \log (\log (2 x))\right )} \, dx-\int \frac {x \log ^2(x)}{(-5+x \log (x)) \log (2 x) \left (-25-11 x+5 x \log (x)+2 x^2 \log (x)+5 \log (\log (2 x))-x \log (x) \log (\log (2 x))\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.09, size = 82, normalized size = 3.57 \begin {gather*} \log \left (25+11 x-5 x (\log (x)-\log (2 x))-2 x^2 (\log (x)-\log (2 x))-5 x \log (2 x)-2 x^2 \log (2 x)-5 \log (\log (2 x))+x (\log (x)-\log (2 x)) \log (\log (2 x))+x \log (2 x) \log (\log (2 x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 + x*Log[x] + (6*x - 2*x^2 + (-5*x - 4*x^2)*Log[x])*Log[2*x] + (x + x*Log[x])*Log[2*x]*Log[Log[2*
x]])/((25*x + 11*x^2 + (-5*x^2 - 2*x^3)*Log[x])*Log[2*x] + (-5*x + x^2*Log[x])*Log[2*x]*Log[Log[2*x]]),x]

[Out]

Log[25 + 11*x - 5*x*(Log[x] - Log[2*x]) - 2*x^2*(Log[x] - Log[2*x]) - 5*x*Log[2*x] - 2*x^2*Log[2*x] - 5*Log[Lo
g[2*x]] + x*(Log[x] - Log[2*x])*Log[Log[2*x]] + x*Log[2*x]*Log[Log[2*x]]]

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fricas [B]  time = 0.71, size = 56, normalized size = 2.43 \begin {gather*} \log \relax (x) + \log \left (-\frac {{\left (2 \, x^{2} + 5 \, x\right )} \log \relax (x) - {\left (x \log \relax (x) - 5\right )} \log \left (\log \relax (2) + \log \relax (x)\right ) - 11 \, x - 25}{x \log \relax (x) - 5}\right ) + \log \left (\frac {x \log \relax (x) - 5}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)+x)*log(2*x)*log(log(2*x))+((-4*x^2-5*x)*log(x)-2*x^2+6*x)*log(2*x)+x*log(x)-5)/((x^2*log(
x)-5*x)*log(2*x)*log(log(2*x))+((-2*x^3-5*x^2)*log(x)+11*x^2+25*x)*log(2*x)),x, algorithm="fricas")

[Out]

log(x) + log(-((2*x^2 + 5*x)*log(x) - (x*log(x) - 5)*log(log(2) + log(x)) - 11*x - 25)/(x*log(x) - 5)) + log((
x*log(x) - 5)/x)

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giac [A]  time = 0.21, size = 37, normalized size = 1.61 \begin {gather*} \log \left (2 \, x^{2} \log \relax (x) - x \log \relax (x) \log \left (\log \relax (2) + \log \relax (x)\right ) + 5 \, x \log \relax (x) - 11 \, x + 5 \, \log \left (\log \relax (2) + \log \relax (x)\right ) - 25\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)+x)*log(2*x)*log(log(2*x))+((-4*x^2-5*x)*log(x)-2*x^2+6*x)*log(2*x)+x*log(x)-5)/((x^2*log(
x)-5*x)*log(2*x)*log(log(2*x))+((-2*x^3-5*x^2)*log(x)+11*x^2+25*x)*log(2*x)),x, algorithm="giac")

[Out]

log(2*x^2*log(x) - x*log(x)*log(log(2) + log(x)) + 5*x*log(x) - 11*x + 5*log(log(2) + log(x)) - 25)

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maple [B]  time = 0.15, size = 48, normalized size = 2.09

method	result	size
risch	\(\ln \relax (x )+\ln \left (\ln \relax (x )-\frac {5}{x}\right )+\ln \left (\ln \left (\ln \relax (2)+\ln \relax (x )\right )-\frac {2 x^{2} \ln \relax (x )+5 x \ln \relax (x )-11 x -25}{x \ln \relax (x )-5}\right )\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*ln(x)+x)*ln(2*x)*ln(ln(2*x))+((-4*x^2-5*x)*ln(x)-2*x^2+6*x)*ln(2*x)+x*ln(x)-5)/((x^2*ln(x)-5*x)*ln(2*x
)*ln(ln(2*x))+((-2*x^3-5*x^2)*ln(x)+11*x^2+25*x)*ln(2*x)),x,method=_RETURNVERBOSE)

[Out]

ln(x)+ln(ln(x)-5/x)+ln(ln(ln(2)+ln(x))-(2*x^2*ln(x)+5*x*ln(x)-11*x-25)/(x*ln(x)-5))

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maxima [B]  time = 0.76, size = 56, normalized size = 2.43 \begin {gather*} \log \relax (x) + \log \left (-\frac {{\left (2 \, x^{2} + 5 \, x\right )} \log \relax (x) - {\left (x \log \relax (x) - 5\right )} \log \left (\log \relax (2) + \log \relax (x)\right ) - 11 \, x - 25}{x \log \relax (x) - 5}\right ) + \log \left (\frac {x \log \relax (x) - 5}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)+x)*log(2*x)*log(log(2*x))+((-4*x^2-5*x)*log(x)-2*x^2+6*x)*log(2*x)+x*log(x)-5)/((x^2*log(
x)-5*x)*log(2*x)*log(log(2*x))+((-2*x^3-5*x^2)*log(x)+11*x^2+25*x)*log(2*x)),x, algorithm="maxima")

[Out]

log(x) + log(-((2*x^2 + 5*x)*log(x) - (x*log(x) - 5)*log(log(2) + log(x)) - 11*x - 25)/(x*log(x) - 5)) + log((
x*log(x) - 5)/x)

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mupad [B]  time = 4.87, size = 57, normalized size = 2.48 \begin {gather*} \ln \left (\frac {x\,\ln \relax (x)-5}{x}\right )+\ln \left (\frac {11\,x-5\,\ln \left (\ln \left (2\,x\right )\right )-2\,x^2\,\ln \relax (x)-5\,x\,\ln \relax (x)+x\,\ln \left (\ln \left (2\,x\right )\right )\,\ln \relax (x)+25}{x\,\ln \relax (x)-5}\right )+\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*log(x) - log(2*x)*(log(x)*(5*x + 4*x^2) - 6*x + 2*x^2) + log(2*x)*log(log(2*x))*(x + x*log(x)) - 5)/(lo
g(2*x)*(25*x - log(x)*(5*x^2 + 2*x^3) + 11*x^2) - log(2*x)*log(log(2*x))*(5*x - x^2*log(x))),x)

[Out]

log((x*log(x) - 5)/x) + log((11*x - 5*log(log(2*x)) - 2*x^2*log(x) - 5*x*log(x) + x*log(log(2*x))*log(x) + 25)
/(x*log(x) - 5)) + log(x)

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sympy [B]  time = 1.66, size = 48, normalized size = 2.09 \begin {gather*} \log {\relax (x )} + \log {\left (\log {\relax (x )} - \frac {5}{x} \right )} + \log {\left (\log {\left (\log {\relax (x )} + \log {\relax (2 )} \right )} + \frac {- 2 x^{2} \log {\relax (x )} - 5 x \log {\relax (x )} + 11 x + 25}{x \log {\relax (x )} - 5} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*ln(x)+x)*ln(2*x)*ln(ln(2*x))+((-4*x**2-5*x)*ln(x)-2*x**2+6*x)*ln(2*x)+x*ln(x)-5)/((x**2*ln(x)-5*
x)*ln(2*x)*ln(ln(2*x))+((-2*x**3-5*x**2)*ln(x)+11*x**2+25*x)*ln(2*x)),x)

[Out]

log(x) + log(log(x) - 5/x) + log(log(log(x) + log(2)) + (-2*x**2*log(x) - 5*x*log(x) + 11*x + 25)/(x*log(x) -
5))

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